
#1
Jul609, 09:35 PM

P: 126

In 1916 Schwarzschild wrote down his famous metric to solve (or resolve using a polar coordinate system) the precession of the perihelion of Mercury. The curvature of spacetime described by the Metric is for any nonrotating spherically symmetric mass.
[tex]ds^2 = (1\frac{2M}{r})dt^2 + (1\frac{2M}{r})^{1}dr^2 + r^2(d\theta^2+sin^2\theta d\phi^2)[/tex] Does this imply there is a black hole at the center of the Sun, the Earth etc? 



#2
Jul609, 11:54 PM

HW Helper
P: 2,156

Nope. That Schwarzschild metric only applies outside a solid body  there's a different metric that describes the inside of the Sun or Earth.




#3
Jul709, 12:14 AM

P: 126

Is the different metric just the Newtonian one?




#4
Jul709, 02:43 AM

P: 3,542

Schwarzschild Metric: The Sun
Interior Schwarzschild solution:
http://www.physicsforums.com/showthr...02#post1543402 



#5
Jul709, 06:03 AM

P: 126

The interior solution according to George Jones:
[tex]d\tau^{2}=\left( \frac{3}{2}\sqrt{1\frac{2M}{R}}\frac{1}{2}\sqrt{1\frac{2Mr^{2}}{R^{3}}}\right) ^{2}dt^{2}\left( 1\frac{2Mr^{2}}{R^{3}}\right) ^{1}dr^{2}r^{2}\left( d\theta ^{2}+\sin ^{2}\theta d\phi ^{2}\right)[/tex] I note that in addition to the singularity at [tex]R=2M[/tex] there is a new singularity at [tex]r^2=R^3[/tex] and the expression goes complex when [tex]r^2 > R^3[/tex] I assume that this means that the interior solution line integral predicts a nasty end for an astronaut at a height of [tex]\sqrt{(6371000^3)}  6371000[/tex] = 16 million kilometers from the surface of the Earth (I'm being facetious) Can I ask how the interior Schwarzschild solution was derived? (Not by Schwarzschild, obviously) 



#6
Jul709, 06:22 AM

Mentor
P: 6,038

General Relativity: An Introduction for Physicists by Hobson, Efstathiou, and Lasenby Gravitation by Misner, Thorne, and Wheeler. [itex]r = R[/itex] represents the surface of the spherical body. If [tex]R > 2M \frac{9}{8},[/tex] then the solution has no singularities. At [tex]R = 2M \frac{9}{8},[/tex] the solution develops a pressure singularity at its centre. This solution is an "almost realistic" toy model for a spherical body. 



#7
Jul709, 06:42 AM

P: 126

George
How can a fluid sphere have a constant density? Isn't that a tad unrealistic? 



#8
Jul709, 07:08 AM

Mentor
P: 6,038

[tex] \frac{dm}{dr} \left( r \right) = 4 \pi r^2 \rho (r) [/tex] and the OppenheimerVolkoff equation (12.21) from the link below. For the case of constant density, this coupled pair of differential equations has an exact solution. For more realistic cases, the existence and uniqueness theorems for differential equations guarantee that a solutions still exist, but, unfortunately, solutions can not be written down in terms of elementary or standard special functions. As is the case for realistic treatments of many areas of physics, numerical methods must be used. See the paragraph on page 293 starting with "The closed system of three equations ..." http://books.google.ca/books?id=xma1...senby&pg=PA293 Even more realistic treatments ditch spherical symmetry (e.g., axisymmetric rotating stars) and consider material other than perfect fluids. 



#9
Jul709, 08:01 AM

P: 126

Thanks George and everyone else who answered. More to think about.




#10
Mar2110, 07:48 PM

P: 479

Sorry for the thread necromancy, but what about the other singularity that diamond pointed out? The r^2 = R^3 (as it happens i recently ran face first into that rather annoying singularity myself so i'm interested as to its properties). Is there a known coordinate transformation that removes it? I can't see any reason why it would be a physical singularity...
Also in the interior metric given, shouldn't the mass term in the radial coefficient be m(r) not M? Cheers G EDIT: Just read the theorem here: http://books.google.ca/books?id=xma1...age&q=&f=false I suppose this answers some questions about the pressure singularity, but thats not the same as the one diamond pointed out right? I mean the physical location of that one depends strongly on if R>1 or R<1 



#11
Mar2210, 05:35 AM

P: 3,966

If you are still interested in discussing the case for non uniform density distribution of M(r) then just ask. I studied this topic a while back. P.S. What we should be concerned with is the case when 2M*r^2>=R^3. I will need to think about that when I am less tired. 



#12
Mar2210, 04:52 PM

P: 3,966

[tex]\frac{d\tau}{dt}= \frac{3}{2}\sqrt{1\frac{2M}{R}}\frac{1}{2}\sqrt{1\frac{2Mr^{2}}{R^{3}}}[/tex] By setting the quantity dtau/dt to zero, I can solve for r to find the location of the singularity and obtain: [tex]r = \pm R \sqrt{9\frac{4R}{M}} [/tex] which is unique if we only consider positive values of r. There are not two separate singularites at R=2M and r^2=R^3 as conjectured by Diamond. Now if we plug in R = (9/8)Rs = 9M/4 the singularity is at r=0 as expected. For R = 2M the singularity has moved out to r = 2M and what is surprising is that for r<2M the ratio of proper time rate to coordinate time rate (dtau/dt) is negative and real . In other words proper time is running in the opposite direction to the coordinate time for a stationary observer inside the the boundary of the massive body. Even more interestingly when R=M (ie the body has collapsed to a physical radius less than the Shwarschild radius) the interior solution predicts there is no singularity or imaginary or complex values for (dtau/dt) according to a stationary observer inside the collapsing body, but the singularity at r=2m remains because the exterior Schwarzschild solution is valid in that region. Now the original equation with average density (d) explicity stated is: [tex]\frac{d\tau}{dt}= \frac{3}{2}\sqrt{1\frac{2M}{R}}\frac{1}{2}\sqrt{1\frac{2M^{2}}{r}\frac{4/3 \pi r^3 d}{4/3 \pi R^3 d}}[/tex] After cancelations it is easy to see the above equation reduces to the familar form at the top of this post. Now if we set density as function d(r) of radius so that d(r) = d/r which is a bit more realistic, then the ratio of the enclosed mass to the total mass of the is proportional to r^2/R^3 rather than r^3/R^3 for the uniform density case and the equation becomes: [tex]\frac{d\tau}{dt}= \frac{3}{2}\sqrt{1\frac{2M}{R}}\frac{1}{2}\sqrt{1\frac{2Mr}{R^{3}}}[/tex] Solving this for r when dtau/dt=0 gives the location of the singularity as: [tex]r = R^2 \left(9\frac{4R}{M}\right) [/tex] Using this slightly more realistic equation (density increasing towards infinite going towards the centre) we can see that a singularity still occurs ar r=0 when R=9/8(2M). 



#13
Mar2210, 07:15 PM

P: 479

Hey kev,
Cheers for the replies! Your point about the reversal of proper time is familiar, i remember reading a paper where someone wanted to redefine the event horizon (or perhaps invent a new one) as the point where this switch occurs. Next question :) Given that the singularity is of pressure, a scalar, you can't do a coordinate transform to get rid of it. So, either I need a new density profile that doesn't give rise to this which means solving the tolmanoppenheimervolkov equation (which i've tried and not fared very well so far but i think that's because i've no idea what i'm doing really) or a different interior solution. In terms of the latter i have found these papers: http://www.jstor.org/stable/78530 http://www.springerlink.com/content/w568765vt481871r/ The second is a comment on the first essentially doing some more checks to make sure its viable, where the first gives a new interior schwarszchild solution that, alledgedly, does not have a pressure singularity even for constant density (the physical viability of this is somewhat dubious to me but anyway). It all seems very good but in the first paper in eqn 1.3 i am unsure as to the difference between R and a. He defines a as the radius of the region bounding the mass, which surely is the radius of the object which is normally R? EDIT: Oops i'm retarded, think i got it now :) Still: would appreciate comments on that metric Cheers G 



#14
Mar2210, 07:40 PM

Sci Advisor
P: 8,470





#15
Mar2210, 09:07 PM

P: 3,966

By considering a stationary event, as in my last post the interior solution in the paper simplifies to: [tex]\frac{d\tau}{dt}= \frac{3}{2}\sqrt{1\frac{a^2}{R_*^2}}\frac{1}{2}\sqrt{1\frac{r^{2}}{R_*^{2}}[/tex] By comparing their equation to the one given by George, it is fairly easy to work out that [tex]a = R[/tex] and [tex]R_*^2 = R^3/2M[/tex]. If we wish to equate [tex]R_*[/tex] with [tex]R[/tex] then the equation is only valid for a massive body has a physical surface ar R=2M and if the observer is located exactly at the event horizon, but that limits the general application of the equation. This is a bit wierd and it would make the clarity of the rest of the article a bit hard to follow. 



#16
Mar2210, 09:19 PM

P: 479

If it isn't and thus as you say i'm then not interested in this solution, what do you suggest for an approach to find an interior solution that does not have this pressure singularity? I've had great trouble finding/solving for one although i understand it can be done numerically but i'd rather not resort to that (I wish to take limits). cheers G 



#17
Mar2210, 09:35 PM

P: 3,966

As far as being valid for external observers you have to superimpose the exterior solution on the interior solution. Work out the clock rate for r<R relative to a clock at 



#18
Mar2210, 10:50 PM

P: 479

Also, sorry, i didn't realise you only had access to the previews. Full text is here: http://members.iinet.net.au/~housewrk/Papers/ Essentially what i want to do is solve the relativistic Klein Gordon equation on the interior of say a star then match that solution to the exterior solution found using the exterior metric then propagate outwards and find a phase shift. I also don't want to be inhibited as to the size/mass of the object hence why that pressure singularity is an issue for me. I have made the substitution you suggested using the new metric he gives in the paper for the constant density profile without the singularity (again, not entirely sure of the validity of this but as i mentionde previously this isn't a coordinate change rather a totally different formulation. I am unsure wether its appropriate or not but am giving it a go just to see what happens) Hope that makes sense! Cheers G 


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